已知函数f(x)=sin^2ωx+√3 sinωxsin(ωx+π/2)(ω>0)的最小正周期为π

1个回答

  • (1)

    f(x) = sin²(ωx) + √3 sin(ωx)sin(ωx+π/2) = sin²(ωx) + √3sin(ωx)cos(ωx)

    = 2[(1/2)sin(ωx) + (√3/2)cos(ωx)]sin(ωx)

    = 2[cos(π/3)sin(ωx) + sin(π/3)cos(ωx)]sin(ωx)

    = 2sin(ωx + π/3)sin(ωx)

    = 2*(1/2)[cos(ωx + π/3 - ωx) - cos(ωx + π/3 + ωx)]

    = -cos(2ωx + π/3) + cos(π/3)

    = -cos(2ωx + π/3) + 1/2

    = cos(2ωx + π/3 - π) + 1/2

    = cos(2ωx - 2π/3) + 1/2

    最小正周期为π = 2π/(2ω),ω = 1

    f(x) = cos(2x - 2π/3) + 1/2

    (2)

    x∈[-π/12,π/2]:

    f(π/3) = cos(2π/3 - 2π/3) + 1/2 = 1 + 1/2 = 3/2,此为最大值

    x = π/3为f(x)最大值处的对称轴

    π/3 - (-π/12) = 5π/12

    π/2 - π/3 = 2π/12 < 5π/12

    x = -π/12比x = π/2与x = π/3对称轴更远,最小值f(-π/12) = cos(-π/6 - 2π/3) + 1/2

    = (1 - √3)/2

    值域:[(1 - √3)/2,3/2]